Problem: Is the function $f(x) = \lfloor x \rfloor + \frac{1}{2}$ even, odd, or neither?

Enter "odd", "even", or "neither".
Since $f \left( \frac{1}{2} \right) = \left\lfloor \frac{1}{2} \right\rfloor + \frac{1}{2} = \frac{1}{2}$ and $f \left( -\frac{1}{2} \right) = \left\lfloor -\frac{1}{2} \right\rfloor + \frac{1}{2} = -\frac{1}{2},$ so if $f$ is either even or odd, it must be odd.

But $f(0) = \lfloor 0 \rfloor + \frac{1}{2}.$  Every odd function $f(x)$ satisfies $f(0) = 0,$ so $f(x)$ is $\boxed{\text{neither}}.$